3.863 \(\int \frac{\left (a+b x^2\right )^2}{(e x)^{5/2} \left (c+d x^2\right )^{5/2}} \, dx\)

Optimal. Leaf size=258 \[ -\frac{\sqrt{e x} \left (3 a^2 d^2-2 a b c d+b^2 c^2\right )}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}-\frac{2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}+\frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (5 a d (2 b c-3 a d)+b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{12 c^{13/4} d^{5/4} e^{5/2} \sqrt{c+d x^2}}+\frac{\sqrt{e x} \left (5 a d (2 b c-3 a d)+b^2 c^2\right )}{6 c^3 d e^3 \sqrt{c+d x^2}} \]

[Out]

(-2*a^2)/(3*c*e*(e*x)^(3/2)*(c + d*x^2)^(3/2)) - ((b^2*c^2 - 2*a*b*c*d + 3*a^2*d
^2)*Sqrt[e*x])/(3*c^2*d*e^3*(c + d*x^2)^(3/2)) + ((b^2*c^2 + 5*a*d*(2*b*c - 3*a*
d))*Sqrt[e*x])/(6*c^3*d*e^3*Sqrt[c + d*x^2]) + ((b^2*c^2 + 5*a*d*(2*b*c - 3*a*d)
)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*Ar
cTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(12*c^(13/4)*d^(5/4)*e^(5/2)*
Sqrt[c + d*x^2])

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Rubi [A]  time = 0.582222, antiderivative size = 258, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179 \[ -\frac{\sqrt{e x} \left (3 a^2 d^2-2 a b c d+b^2 c^2\right )}{3 c^2 d e^3 \left (c+d x^2\right )^{3/2}}-\frac{2 a^2}{3 c e (e x)^{3/2} \left (c+d x^2\right )^{3/2}}+\frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (5 a d (2 b c-3 a d)+b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{12 c^{13/4} d^{5/4} e^{5/2} \sqrt{c+d x^2}}+\frac{\sqrt{e x} \left (5 a d (2 b c-3 a d)+b^2 c^2\right )}{6 c^3 d e^3 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]  Int[(a + b*x^2)^2/((e*x)^(5/2)*(c + d*x^2)^(5/2)),x]

[Out]

(-2*a^2)/(3*c*e*(e*x)^(3/2)*(c + d*x^2)^(3/2)) - ((b^2*c^2 - 2*a*b*c*d + 3*a^2*d
^2)*Sqrt[e*x])/(3*c^2*d*e^3*(c + d*x^2)^(3/2)) + ((b^2*c^2 + 5*a*d*(2*b*c - 3*a*
d))*Sqrt[e*x])/(6*c^3*d*e^3*Sqrt[c + d*x^2]) + ((b^2*c^2 + 5*a*d*(2*b*c - 3*a*d)
)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*Ar
cTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(12*c^(13/4)*d^(5/4)*e^(5/2)*
Sqrt[c + d*x^2])

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Rubi in Sympy [A]  time = 56.4105, size = 236, normalized size = 0.91 \[ - \frac{2 a^{2}}{3 c e \left (e x\right )^{\frac{3}{2}} \left (c + d x^{2}\right )^{\frac{3}{2}}} - \frac{\sqrt{e x} \left (a d \left (3 a d - 2 b c\right ) + b^{2} c^{2}\right )}{3 c^{2} d e^{3} \left (c + d x^{2}\right )^{\frac{3}{2}}} + \frac{\sqrt{e x} \left (- 5 a d \left (3 a d - 2 b c\right ) + b^{2} c^{2}\right )}{6 c^{3} d e^{3} \sqrt{c + d x^{2}}} + \frac{\sqrt{\frac{c + d x^{2}}{\left (\sqrt{c} + \sqrt{d} x\right )^{2}}} \left (\sqrt{c} + \sqrt{d} x\right ) \left (- 5 a d \left (3 a d - 2 b c\right ) + b^{2} c^{2}\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}} \right )}\middle | \frac{1}{2}\right )}{12 c^{\frac{13}{4}} d^{\frac{5}{4}} e^{\frac{5}{2}} \sqrt{c + d x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((b*x**2+a)**2/(e*x)**(5/2)/(d*x**2+c)**(5/2),x)

[Out]

-2*a**2/(3*c*e*(e*x)**(3/2)*(c + d*x**2)**(3/2)) - sqrt(e*x)*(a*d*(3*a*d - 2*b*c
) + b**2*c**2)/(3*c**2*d*e**3*(c + d*x**2)**(3/2)) + sqrt(e*x)*(-5*a*d*(3*a*d -
2*b*c) + b**2*c**2)/(6*c**3*d*e**3*sqrt(c + d*x**2)) + sqrt((c + d*x**2)/(sqrt(c
) + sqrt(d)*x)**2)*(sqrt(c) + sqrt(d)*x)*(-5*a*d*(3*a*d - 2*b*c) + b**2*c**2)*el
liptic_f(2*atan(d**(1/4)*sqrt(e*x)/(c**(1/4)*sqrt(e))), 1/2)/(12*c**(13/4)*d**(5
/4)*e**(5/2)*sqrt(c + d*x**2))

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Mathematica [C]  time = 0.448213, size = 211, normalized size = 0.82 \[ \frac{x^{5/2} \left (\frac{a^2 (-d) \left (4 c^2+21 c d x^2+15 d^2 x^4\right )+2 a b c d x^2 \left (7 c+5 d x^2\right )+b^2 c^2 x^2 \left (d x^2-c\right )}{c^3 d x^{3/2} \left (c+d x^2\right )}+\frac{i x \sqrt{\frac{c}{d x^2}+1} \left (-15 a^2 d^2+10 a b c d+b^2 c^2\right ) F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}{\sqrt{x}}\right )\right |-1\right )}{c^3 d \sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}\right )}{6 (e x)^{5/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]  Integrate[(a + b*x^2)^2/((e*x)^(5/2)*(c + d*x^2)^(5/2)),x]

[Out]

(x^(5/2)*((b^2*c^2*x^2*(-c + d*x^2) + 2*a*b*c*d*x^2*(7*c + 5*d*x^2) - a^2*d*(4*c
^2 + 21*c*d*x^2 + 15*d^2*x^4))/(c^3*d*x^(3/2)*(c + d*x^2)) + (I*(b^2*c^2 + 10*a*
b*c*d - 15*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/S
qrt[d]]/Sqrt[x]], -1])/(c^3*Sqrt[(I*Sqrt[c])/Sqrt[d]]*d)))/(6*(e*x)^(5/2)*Sqrt[c
+ d*x^2])

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Maple [B]  time = 0.037, size = 686, normalized size = 2.7 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((b*x^2+a)^2/(e*x)^(5/2)/(d*x^2+c)^(5/2),x)

[Out]

-1/12*(15*(-c*d)^(1/2)*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-
c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)
^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^3*a^2*d^3-10*(-c*d)^(1/2)*((d*x+(-c*d
)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-
x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^
(1/2))*x^3*a*b*c*d^2-(-c*d)^(1/2)*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2
)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF((
(d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^3*b^2*c^2*d+15*(-c*d)^(1/2
)*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1
/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))
^(1/2),1/2*2^(1/2))*x*a^2*c*d^2-10*(-c*d)^(1/2)*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2)
)^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/
2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x*a*b*c^2*d-(-
c*d)^(1/2)*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/
(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*
d)^(1/2))^(1/2),1/2*2^(1/2))*x*b^2*c^3+30*x^4*a^2*d^4-20*x^4*a*b*c*d^3-2*x^4*b^2
*c^2*d^2+42*x^2*a^2*c*d^3-28*x^2*a*b*c^2*d^2+2*x^2*b^2*c^3*d+8*a^2*c^2*d^2)/x/e^
2/(e*x)^(1/2)/c^3/d^2/(d*x^2+c)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (b x^{2} + a\right )}^{2}}{{\left (d x^{2} + c\right )}^{\frac{5}{2}} \left (e x\right )^{\frac{5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^2 + a)^2/((d*x^2 + c)^(5/2)*(e*x)^(5/2)),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2/((d*x^2 + c)^(5/2)*(e*x)^(5/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}{{\left (d^{2} e^{2} x^{6} + 2 \, c d e^{2} x^{4} + c^{2} e^{2} x^{2}\right )} \sqrt{d x^{2} + c} \sqrt{e x}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^2 + a)^2/((d*x^2 + c)^(5/2)*(e*x)^(5/2)),x, algorithm="fricas")

[Out]

integral((b^2*x^4 + 2*a*b*x^2 + a^2)/((d^2*e^2*x^6 + 2*c*d*e^2*x^4 + c^2*e^2*x^2
)*sqrt(d*x^2 + c)*sqrt(e*x)), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x**2+a)**2/(e*x)**(5/2)/(d*x**2+c)**(5/2),x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (b x^{2} + a\right )}^{2}}{{\left (d x^{2} + c\right )}^{\frac{5}{2}} \left (e x\right )^{\frac{5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^2 + a)^2/((d*x^2 + c)^(5/2)*(e*x)^(5/2)),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2/((d*x^2 + c)^(5/2)*(e*x)^(5/2)), x)